$ABCD$ is a square. $M$ and $N$ are points on $AB $ and $BC$, respectively such that $\angle MDN=45^\circ$...

$ABCD$ is a square. $M$ and $N$ are points on $AB $ and $BC$, respectively such that $\angle MDN=45^\circ$. $R$ is the midpoint of $MN$ and $P$ and $Q$ are the points where $AC$ is intersected by $DM$ and $DN$, respectively. Show that $PR=QR$.

Consider a $(D;\vec{DA},\vec{DC})$ as a coordinate system, then the coordinates of $M$ and $N$ would be $(1,m)$ and $(n,1)$ where $m$ and $n$ are related by the fact that $\angle MDN=45^\circ$. That is $$n=\tan\angle NDC=\tan\left(\frac\pi4-\angle MDA\right)=\frac{1-m}{1+m}\tag1$$ Now the line $DM$ of equation $y=mx$ intersects $AC$ of equation $x+y=1$ at $P\left(\frac{1}{1+m},\frac{m}{1+m}\right)$.

Similarly, the line $DN$ of equation $x=ny$ intersects $AC$ of equation $x+y=1$ at $Q\left(\frac{n}{1+n},\frac{1}{1+n}\right)$. Or according to $(1)$ : $Q\left(\frac{1-m}{2},\frac{1+m}{2}\right)$.

Moreover, the midpoint of $MN$ is $R\left(\frac{1+n}{2},\frac{1+m}{2}\right)$, or $R\left(\frac{1}{1+m},\frac{1+m}{2}\right)$, using $(1)$ once again.

Finally, It is straightforward now to check that $$ \vec{QR}=\left(\frac{1+m^2}{2(1+m)},0\right),\quad \vec{PR}=\left(0,\frac{1+m^2}{2(1+m)}\right) $$ It follows that $QR=PR$ and that $QR\bot PR$.

A Geometric Solution:

$DCNP$ is a cyclic quadrilateral, since $\angle NCP=\angle NDP=\dfrac{\pi}{4}$. Thus, $\angle DPN+\angle DCN=\pi$, that is $\angle DPN=\dfrac{\pi}{2}$. It follows that $PR$ is the median of the right triangle $NPM$ at $P$. In Particular, $PR=\frac{1}{2}MN$.

In a similar way we prove that $QR=\frac{1}{2}MN$ (interchanging the roles of ($A$, $P$, $M$) and ($B$,$Q$, $N$)). And the desired conclusion follows.

Let $S$ be the orthogonal projection of $D$ on $MN$, then $NCDS$ is cyclic. Also, from $\angle PCN=\angle ACB=45^\circ =\angle MDN=\angle PDN$, we have that $NCDP$ is cyclic. Hence, $NCDPS$ is cyclic, in particular, $\angle MPN=\angle DCN=90^\circ$. Analogously we get $\angle MQN=90^\circ$. Hence, $MNQP$ is cyclic and $MN$ is the diameter of $(MNQP)$, therefore, $R$ is the center of $(MNQP)$, this implies that $PR=QR$.